Starting with Rust programming, simple UAL

Last update: Dec 10, 2022

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Overview

AVR-like Instruction set simulator

Starting with Rust programming, simple UAL.

Concepts learned:

ownership
structure
patern matching
data collection
error handling
testing
closures
iterators
conditions ...

On-going development(it's my First Rust program so I `for sure` have made mistakes.

Can we run DOOM on it?

fn main() {
    println!("Sure, DOOM runs on everything :)");
}

TODOS

Add conditional jump ( branche if not equal, jump, ...)
Store multiple u8 data in the u32 memory space, not a single u8
Add logical and arithmetical operation
Use the last 1/2 byte of the instruction in order to manipulate values > 0xff
Better parse system to read instruction like this :

    00000000 <app_name>:
        0:       1120           add      r0,     r1
        1:       3420           mov      r3,     r1
        2:       1120           add      r0,     r1
        3:       1120           add      r0,     r1
        ...           ...           ...         ...
        
        ff:      1120           add      r0,     r1

RUN

cargo run application/prog.S

prog.S (note `\n` at the end of each line, including _start and _end lines)

_start
0x8190; ldi r0, 0x09
0xBB10; st  0x0B, r0
0x92B0; lds r1, 0x0B
_end

std output

AVR-like Instruction Set simulator
Memory Map 
Address (u8) 		 Value (u32) 
9 	 => 		 0
15 	 => 		 0
8 	 => 		 0
6 	 => 		 37552
1 	 => 		 0
3 	 => 		 0
11 	 => 		 9
12 	 => 		 0
13 	 => 		 0
2 	 => 		 0
7 	 => 		 0
0 	 => 		 0
10 	 => 		 0
14 	 => 		 0
5 	 => 		 47888
4 	 => 		 33168
Registers 
Register name) 		 Value (u8) 
r9 	 => 		 0
r1 	 => 		 9
LR 	 => 		 0
r11 	 => 		 0
PSR 	 => 		 0
r6 	 => 		 0
r10 	 => 		 0
r7 	 => 		 0
r2 	 => 		 0
r12 	 => 		 0
r3 	 => 		 0
r0 	 => 		 9
r5 	 => 		 0
r8 	 => 		 0
SP 	 => 		 0
r4 	 => 		 0

Memory dump after loading a program 1

test program

_start
     0  0x8190; ldi r0, 0x09
     1  0xBB10; st  0x0B, r0
     2  0x92B0; lds r1, 0x0B
_end

Memory Map and register Map (register r0 and r1 have their new content

Memory Map 
Address (u8) 		 Value (u32) 
2 	 => 		 0
0 	 => 		 0
1 	 => 		 0
13 	 => 		 0
7 	 => 		 0
5 	 => 		 47888
12 	 => 		 0
3 	 => 		 0
4 	 => 		 33168
11 	 => 		 9
14 	 => 		 0
8 	 => 		 0
9 	 => 		 0
6 	 => 		 37552
10 	 => 		 0
15 	 => 		 0
Registers 
Register name) 		 Value (u8) 
SP 	 => 		 0
r4 	 => 		 0
r10 	 => 		 0
r11 	 => 		 0
r9 	 => 		 0
r3 	 => 		 0
r1 	 => 		 9
r2 	 => 		 0
r7 	 => 		 0
LR 	 => 		 0
PSR 	 => 		 0
r5 	 => 		 0
r6 	 => 		 0
r12 	 => 		 0
r8 	 => 		 0
r0 	 => 		 9

Memory dump after loading a program 2

test program

_start
     0   0x1120; add r0, r1
     1   0xA510; mov r4, r0
     2   0x8150; ldi r0, 5
     3   0x9210; lds r1, 0x01
     4   0x4520; and r4, r1
     5   0x3510; or r4, r0
_end

Memory Map

Program start at address 0x04 in memory, so 0x04, 0x05, 0x06, 0x07, 0x08, 0x09 are modified with their respective machine code.

Memory Map 
Address (u8) 		        Value (u32) 
9 	        => 		 13584
10 	        => 		 0
2 	        => 		 0
13 	        => 		 0
6 	        => 		 33104
14 	        => 		 0
5 	        => 		 42256
3 	        => 		 0
15 	        => 		 0
1 	        => 		 0
0 	        => 		 0
12 	        => 		 0
4 	        => 		 4384
11 	        => 		 0
7 	        => 		 37392
8 	        => 		 17696

Processor architecture

16 register from r0 to r15
16 memory blocks divided into program and data each block has 32 bits total 512 bits or 64 bytes

Register (16)

Register	addr	Register	addr
r0	0x01	r8	0x09
r1	0x02	r9	0x0A
r2	0x03	r10	0x0B
r3	0x04	r11	0x0C
r4	0x05	r12	0x0D
r5	0x06	PSR	0x0E
r6	0x07	SP	0x0F
r7	0x08	LR	0x10

(to add more)

Instruction Set :: Control (0, 5, 6)

op	operands	binary	operands	comment
nop	--	0000	0000 0000 0000	nop operation, cpu does onothing
push	sp?	0101	0000 0000 0000	push to the stack pointer
pop	sp?	0110	0000 0000 0000	pop the stack pointer

Instruction Set :: Logic (1-4)

op	operands	binary	operands	comment
add	rd, rs	0001	rd rd rd rd rs rs rs rs 0000	add rs to rd and return rd
sub	rd ,rs	0010	rd rd rd rd rs rs rs rs 0000	substract rs to rd and return rd
or	rd, rs	0011	rd rd rd rd rs rs rs rs 0000	logical or, rs to rd and return rd
and	rd, rs	0100	rd rd rd rd rs rs rs rs 0000	logical and, rs to rd and return rd

Instruction Set :: Change flow (7)

op	operands	binary	operands	comment
jmp	addr	0111	adr adr adr adr adr adr adr adr 0000	jump to address addr

Instruction Set :: Data transfer (8-10)

op	operands	binary	operands	comment
ldi	rd, K	1000	rd rd rd rd K K K K 0000	Load Immediate the literal K to register rd
lds	rd, k	1001	rd rd rd rd k k k k 0000	Load Direct from Data Space at addresse k to register rd
mov	rd, rs	1010	rd rd rd rd rs rs rs rs 0000	move rs to rd and return rd
st	k, rs	1011	k k k k rs rs rs rs 0000	Sore direct the content of rs into the memory address k

Memory Map (512 bits)

Addresse (1 byte)	Value (4 bytes)	Addresse (1 byte)	Value (4 bytes)
0x00	0x0000_0000	0x08	0x0000_0000
0x01	0x0000_0000	0x09	0x0000_0000
0x02	0x0000_0000	0x0A	0x0000_0000
0x03	0x0000_0000	0x0B	0x0000_0000
0x04	0x0000_0000	0x0C	0x0000_0000
0x05	0x0000_0000	0x0D	0x0000_0000
0x06	0x0000_0000	0x0E	0x0000_0000
0x07	0x0000_0000	0x0F	0x0000_0000

Example of Instruction execution

r2 is register 0011, r3 is register 0100, remaining unused

1010 0011 0100 0000; machine code

;same as 
    0   0xA340 ; MOV r2, r3 move content of r3 into r2

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